Joe Armstrong (AL/EAB)
Wed Jan 25 09:03:02 CET 2006
1 year = 3.15 *10^^7 seconds
Let's buy a fast machine say with a 4GHz clock
This is 4*10^9 * 3.15*10^7 Hz/year = 1.26 * 10^17
Let's assume we can do one computation in 1 instructions
(a very silly assumption)
Then on one machine the computation will take
3.3*10^19/1.26*10^17 years = 2.6*10^2 years
But I very much doubt if you can do one computation in
1 machine cycle - suppose it takes 10^5 cycles (just a guess)
Then it will take 2.6*10^7 years -
That's on one machine.
To do this in Erlang you need to:
1) Write a peer 2 peer infrastructure for performing massive
2) Write a really snazzy application to get millions of people
to install the p2p instrastructure
3) Use the idle time on all the hosts in the p2p application to
do your computation
If you can get 10^8 hosts running you program (it will then be more popular than Skype)
Then you might, with a bit of luck, get an answer.
> -----Original Message-----
> [mailto:] On Behalf Of Nils Müllner
> Sent: den 24 januari 2006 11:31
> Subject: Newbie Question
> I', new to Erlang.
> My Task: I have Key, Vector and Plaintext given for
> crypto:aes_cbc_128_encrypt(Key, Vector and Plaintext). Now I
> need to implement some kind of Brute Force to test all Keys
> and Vectors till i find the right ones for a direct match
> (compare plaintext /w result) and measure time.
> What is the best way to travers through the keys? Should i
> make lists and travers through them? What loop is best for
> performance for about
> 3.3*10^19 iterations? Is there already a
> brute-force-algorithm in erlang?
> Kind regards,
> Nils Muellner
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